How can I prove that the additive group of rationals is not isomorphic to a direct product of two nontrivial groups?

I am working through Paolo Aluffi's new GSM text on my own (self-study). On page 63, he asks the reader to

Prove that $\Bbb{Q}$ is not the direct product of two nontrivial groups.

For some context, this is an exercise following a section entitled "The category Grp". I am assuming that he means "is not isomorphic to the direct product of two nontrivial groups", and I can see two possible ways to proceed with this proof, but have been unsuccessful with either approach.

Approach 1: Show that the additive group of rationals has a property that is preserved by isomorphism that the direct product of two nontrivial groups does not have or vice-versa. This seems challenging unless I can significantly narrow down the properties that a direct product of two nontrivial groups that was isomorphic to $\Bbb{Q}$ would necessarily have.

Approach 2: Considering the section in which this question occurs, show that if $G$ and $H$ are nontrivial groups and $\Bbb{Q} \cong G \times H$, then there are homomorphisms $\varphi_{G}:\Bbb{Q} \rightarrow G$ and $\varphi_{H}:\Bbb{Q} \rightarrow H$ which do not factor or do not factor uniquely through the product $G \times H$. This would be a contradiction, as $G \times H$ is a final object in the category Grp.

I would greatly appreciate suggestions on how to proceed further with either of these approaches or with alternate approaches.

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For any $p,q\in \mathbb Q\setminus 0$, there are non-zero integers $n$ and $m$ such that $np = mq\neq 0$. That is not true for the product of two non-trivial groups. –  Thomas Andrews Dec 19 '11 at 22:02
Is there more than one David Pincus, or do I have two of your papers sitting on my desk? –  Asaf Karagila Dec 19 '11 at 22:05
IOW Thomas' hint goes together with your Approach #1: A direct product has a pair of non-trivial subgroups that intersects trivially whereas –  Jyrki Lahtonen Dec 19 '11 at 22:05
Thank you Thomas, Jyrki, and François; your hints were more than sufficient. A direct product $G \times H$ of nontrivial groups has a pair of subgroups $(G \times \{e_H\}$ and $\{e_G\} \times H)$ that intersect trivially. An isomorphism from $G \times H$ onto $\Bbb{Q}$ would map each of these subgroups onto nontrivial subgroups of $\Bbb{Q}$ that intersect trivially. But as you all pointed out any two nontrivial subgroups of $\Bbb{Q}$ intersect nontrivially. Q.E.D. –  David Pincus Dec 20 '11 at 2:51
Isomorphisms are, by definition, homomorphism that have an inverse that is also a homomorphism. So, yes, isomorphism is, by definition, assumed to be a homomorphism. –  Arturo Magidin Mar 30 '12 at 18:20
What is the endomorphism ring isomorphic to? To itself? I thought about the fact that $\mathbb Q$ cannot be isomorphic to a direct product of two rings (because one has non-trivial ideals and $\mathbb Q$ does not), but didn't know how to use that property. –  Patrick Da Silva Dec 19 '11 at 22:11
@PatrickDaSilva, indeed, the endomorphism ring of the abelian group $\mathbb Q$ is isomorphic to ring $\mathbb Q$. –  Mariano Suárez-Alvarez Dec 19 '11 at 22:13
@Patrick: The simple way to establish this is to show that if you know what $1$ maps to, then you know where everything maps to, and that $1$ can map anywhere. $\mathbb{Q}$ is a "free torsionfree divisible group of rank 1". –  Arturo Magidin Mar 30 '12 at 18:22